bzoj 1702: [Usaco2007 Mar]Gold Balanced Lineup 平衡的队列

Description

Farmer John's N cows (1 <= N <= 100,000) share many similarities. In fact, FJ has been able to narrow down the list of features shared by his cows to a list of only K different features (1 <= K <= 30). For example, cows exhibiting feature #1 might have spots, cows exhibiting feature #2 might prefer C to Pascal, and so on. FJ has even devised a concise way to describe each cow in terms of its "feature ID", a single K-bit integer whose binary representation tells us the set of features exhibited by the cow. As an example, suppose a cow has feature ID = 13. Since 13 written in binary is 1101, this means our cow exhibits features 1, 3, and 4 (reading right to left), but not feature 2. More generally, we find a 1 in the 2^(i-1) place if a cow exhibits feature i. Always the sensitive fellow, FJ lined up cows 1..N in a long row and noticed that certain ranges of cows are somewhat "balanced" in terms of the features the exhibit. A contiguous range of cows i..j is balanced if each of the K possible features is exhibited by the same number of cows in the range. FJ is curious as to the size of the largest balanced range of cows. See if you can determine it.

N(1<=N<=100000)头牛，一共K(1<=K<=30)种特色，
每头牛有多种特色，用二进制01表示它的特色ID。比如特色ID为13(1101)，
则它有第1、3、4种特色。[i,j]段被称为balanced当且仅当K种特色在[i,j]内
拥有次数相同。求最大的[i,j]段长度。

Input

* Line 1: Two space-separated integers, N and K.

* Lines 2..N+1: Line i+1 contains a single K-bit integer specifying the features present in cow i. The least-significant bit of this integer is 1 if the cow exhibits feature #1, and the most-significant bit is 1 if the cow exhibits feature #K.

Output

* Line 1: A single integer giving the size of the largest contiguous balanced group of cows.

Sample Input

7 3
7
6
7
2
1
4
2

INPUT DETAILS:

The line has 7 cows with 3 features; the table below summarizes the
correspondence:
Feature 3: 1 1 1 0 0 1 0
Feature 2: 1 1 1 1 0 0 1
Feature 1: 1 0 1 0 1 0 0
Key: 7 6 7 2 1 4 2
Cow #: 1 2 3 4 5 6 7

Sample Output

4

OUTPUT DETAILS:

In the range from cow #3 to cow #6 (of size 4), each feature appears
in exactly 2 cows in this range:
Feature 3: 1 0 0 1 -> two total
Feature 2: 1 1 0 0 -> two total
Feature 1: 1 0 1 0 -> two total
Key: 7 2 1 4
Cow #: 3 4 5 6

类似前缀和。每位分别做前缀和。然后减去第一位的数。每位均相同则表示区间可取。

http://blog.csdn.net/hqd_acm/article/details/5902792

#include
#include
#include
#include
using namespace std;
int hash[100007][34];
int mod=100007;
int a[100001][31];
int s[100001][31];
int k;
inline bool check(int t,int xt)
{
     int i;
     bool flag=true;
     for(i=0;i<=k-1;i++)
          if(s[xt][i]!=hash[t][i])
               return false;
     return true;
}
inline int find(int x,int xt,int xp)
{
	 int t=x;
     while(hash[t][32]!=-1)
     {
	      if(!check(t,xt))
               t=(t+1)%mod;
          else
               break;
     }
     if(hash[t][32]==-1)
     {
          int i;
          for(i=0;i<=k-1;i++)
               hash[t][i]=s[xt][i];
          hash[t][33]=xp;
          hash[t][32]=1;
          return xp;
     }
     return hash[t][33];
}
int main()
{
     int n;
     scanf("%d%d",&n,&k);
     int i,j;
     int x;
     for(i=1;i<=n;i++)
     {
          scanf("%d",&x);
          int p=0;
          while(x!=0)
          {
          	   a[i][p]=x%2;
               x=x/2;
               p++;
          }
     }
     for(i=1;i<=n;i++)
          for(j=0;j<=k-1;j++)
               s[i][j]=s[i-1][j]+a[i][j];
   /*  for(i=1;i<=n;i++)
     {
          for(j=k-1;j>=0;j--)
               printf("%d",s[i][j]);
          printf("\n");
     }
     printf("\n");*/
     for(i=1;i<=n;i++)
          for(j=k-1;j>=0;j--)
               s[i][j]-=s[i][0];
     memset(hash,-1,sizeof(hash));
     /*for(i=1;i<=n;i++)
     {
          for(j=k-1;j>=0;j--)
               printf("%d",s[i][j]);
          printf("\n");
     }
     printf("\n");*/
     int ans=0;
     for(i=0;i<=n;i++)
     {
     	  int p=0;
          for(j=k-1;j>=0;j--)
          {
               p=(p*4+s[i][j])%mod;
               while(p<0)
                    p=-p;
          }
         /* int p=0;
          for(j=0;j<=k-1;j++)
		       p=((p<<2)+(s[i][j]>>4))^(s[i][j]<<10);
          p=p%mod;
          if(p<0) 
		       p=p+mod;*/
        //  printf("%d ",p);
          int loc=find(p,i,i);
          ans=max(ans,i-loc);
	 }
	// printf("\n");
	 printf("%d\n",ans);
     return 0;
}